Earth can be thought of as a sphere of radius $6400 \mathrm{km}$. Any object (or a person) is performing circular motion around the axis of the earth due to earth's rotation (period $1$ day). What is the acceleration of object on the surface of earth (at equator) towards its centre? What is it at latitude $\theta$ ? How does these accelerations compare with $\mathrm{g}=9.8 \mathrm{m} {\mathrm{s}}^{-2}$?

If $\left|A\right|=2$ and $\left|B\right|=4$ then match the relations in column I with the angle $\theta$ between $A$ and $B$ is column II.

A hill is $500 \mathrm{m}$ high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of $125 \mathrm{m} {\mathrm{s}}^{-1}$ over the hill. The canon is located at a distance of $800 \mathrm{m}$ from the foot of the hill and can be moved on the ground at a speed of $2 \mathrm{m} {\mathrm{s}}^{-1}$; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take $g=10 \mathrm{m} {\mathrm{s}}^{-2}$.

A gun can fire shells with maximum speed $v_0$ and the maximum horizontal range that can be achieved is $R=\frac{v_0^2}{g}$.If a target farther away by distance $\Delta x$ (beyond $R$ ) has to be hit with the same gun as shown in the figure here. Show that it could be achieved by raising the gun to a height at least $\mathrm{h}=\Delta x\left[1+\frac{\Delta x}{R}\right]$.

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle with speed $v_0$ and rebounds elastically as shown in the figure. Find the distance along the plane where it will hit the second time.

A girl riding a bicycle with a speed of $5 \mathrm{m} {\mathrm{s}}^{-1}$ towards north direction, observes rain falling vertically down. If she increases her speed to$10 \mathrm{m} {\mathrm{s}}^{-1}$, rain appears to meet her at $45°$ to the vertical. What is the speed of the rain? In what direction does rainfall as observed by a ground-based observer? 

A man wants to reach from $\mathrm{A}$ to the opposite corner of the square $\mathrm{C}$ (as in figure) The sides of the square are $100 \mathrm{m}$. A central square of $50 \mathrm{m}\times 50 \mathrm{m}$ is filled with sand. Outside this square, he can walk at a speed $1\mathrm{m} {\mathrm{s}}^{-1}$. In the central square, he can walk only at a speed of $v \mathrm{m} {\mathrm{s}}^{-1}(v<1)$. What is smallest value of $v$ for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

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